Penyisihan LMNas UGM SMA Ke-28

\(\begin{aligned} \sin2a&=2\sin a\cos a\\ \cos a&=\dfrac{\sin 2a}{2\sin a} \end{aligned}\)

\(\begin{aligned} \cos\left(56°\right)\cdot\cos\left(2\cdot56°\right)\cdot\cos\left(2^2\cdot56°\right)\cdots\cos\left(2^{23}\cdot56°\right)&=\dfrac{\sin\left(2\cdot56°\right)}{2\sin\left(56°\right)}\cdot\dfrac{\sin\left(2^2\cdot56°\right)}{2\sin\left(2\cdot56°\right)}\cdot\dfrac{\sin\left(2^3\cdot56°\right)}{2\sin\left(2^2\cdot56°\right)}\cdots\dfrac{\sin\left(2^{24}\cdot56°\right)}{2\sin\left(2^{23}\cdot56°\right)}\\[4pt] &=\dfrac{\sin\left(2^{24}\cdot56°\right)}{2^{24}\sin\left(56°\right)} \end{aligned}\)
Kita buktikan bahwa 2²⁴⋅56° = 56° + k⋅360°, atau 2²⁴⋅56 = 56 mod 360.
\(\begin{aligned} 2^{24}\cdot56\mod360&\equiv8\left(2^{24}\cdot7\mod45\right)\mod360\\ &\equiv8\left(7\mod45\right)\mod360&\ {\color{red}\phi(45)=24}\\ &\equiv56\mod360 \end{aligned}\)

\(\begin{aligned} \dfrac{\sin\left(2^{24}\cdot56°\right)}{2^{24}\sin\left(56°\right)}&=\dfrac{\sin\left(56°\right)}{2^{24}\sin\left(56°\right)}\\[4pt] &=\dfrac1{2^{24}}\\ &=\color{blue}\boxed{\boxed{\color{black}2^{-24}}} \end{aligned}\)

Untuk x = 28,
\(\begin{aligned} 28P(27)&=(28-28)P(28)\\ P(27)&=0 \end{aligned}\)

Karena $P(x-1)=\dfrac{x-28}xP(x)$, maka P(27) = P(26) = ⋯ = P(1) = P(0) = 0. Sehingga,
P(x) = ax (x − 1)(x − 2)⋯(x − 27)
Agar P(28) = 28!, maka a = 1.
Hanya ada 1 polinom.

\(\begin{aligned} \left(x+\dfrac1x\right)^2&=3\\[4pt] x^2+\dfrac1{x^2}+2&=3\\[4pt] x^2+\dfrac1{x^2}&=1\\[4pt] x^4+1&=x^2\\ x^4&=x^2-1\\ x^6&=x^4-x^2\\ x^6&=-1 \end{aligned}\)

\(\begin{aligned} x^{2017}+x^{67}+x^{44}+x^{31}+x^{26}+x+6&=\left(x^6\right)^{336}\cdot x+\left(x^6\right)^{11}\cdot x+\left(x^6\right)^7\cdot x^2+\left(x^6\right)^5\cdot x+\left(x^6\right)^4\cdot x^2+x+6\\ &=\left(-1\right)^{336}\cdot x+\left(-1\right)^{11}\cdot x+\left(-1\right)^7\cdot x^2+\left(-1\right)^5\cdot x+\left(-1\right)^4\cdot x^2+x+6\\ &=x-x-x^2-x+x^2+x+6\\ &=\color{blue}\boxed{\boxed{\color{black}6}} \end{aligned}\)

\(\begin{aligned} \sin^4t-\cos^4t&=1\\ \left(\sin^2t+\cos^2t\right)\left(\sin^2t-\cos^2t\right)&=1\\ \sin^2t-\cos^2t&=1\\ \cos^2t-\sin^2t&=-1\\ \cos2t&=-1\\ \cos2t&=\cos\pi \end{aligned}\)

• 2t = π + 2kπ
  $t=\dfrac{\pi}2+k\pi$
  
  k = −1 ⟶ $t=-\dfrac{\pi}2$ k = 0 ⟶ $t=\dfrac{\pi}2$
• 2t = −π + 2kπ
  $t=-\dfrac{\pi}2+k\pi$
  
  k = 0 ⟶ $t=-\dfrac{\pi}2$ k = 1 ⟶ $t=\dfrac{\pi}2$

Misal k₁, k₂, ..., k₂₀₁₇ > 0 merupakan akar-akar dari p(x).
k₁k₂⋯k₂₀₁₇ = 2017

p(x) = a(x − k₁)(x − k₂)⋯(x − k₂₀₁₇)

\(\begin{aligned} q(x)&=p\left(x^2\right)\\ &=a\left(x^2-k_1\right)\left(x^2-k_2\right)\cdots\left(x^2-k_{2017}\right)\\ &=a\left(x+\sqrt{k_1}\right)\left(x-\sqrt{k_1}\right)\left(x+\sqrt{k_2}\right)\left(x-\sqrt{k_2}\right)\cdots\left(x+\sqrt{k_{2017}}\right)\left(x-\sqrt{k_{2017}}\right)\\ \end{aligned}\)
Akar-akar dari q(x) adalah
$-\sqrt{k_1},-\sqrt{k_2},...,-\sqrt{k_{2017}},\sqrt{k_1},\sqrt{k_2},...,\sqrt{k_{2017}}$

\(\begin{aligned} b_1b_2\cdots b_{2017}&=\left(-\sqrt{k_1}\right)\left(-\sqrt{k_2}\right)\cdots\left(-\sqrt{k_{2017}}\right)\\ &=-\sqrt{k_1k_2\cdots k_{2017}}\\ &=\color{blue}\boxed{\boxed{\color{black}-\sqrt{2017}}} \end{aligned}\)

Perhatikan bahwa
n mod 9 ≡ S(n) mod 9
n − S(n) mod 9 ≡ 0 mod 9
f(n) mod 9 ≡ 0 mod 9
$\underbrace{(f\circ f\circ f \circ \cdots \circ f)}_{sebanyak\ 2017}\left(2017^{2017}\right)\mod9\equiv0\mod9$

\(\begin{aligned} \log_ab&=\dfrac32\\[4pt] a^3&=b^2=m^6 \end{aligned}\)
a = m², b = m³

\(\begin{aligned} \log_cd&=\dfrac54\\[4pt] c^5&=d^4=n^{20} \end{aligned}\)
c = n⁴, d = n⁵

\(\begin{aligned} a-c&=9\\ m^2-n^4&=9\\ \left(m+n^2\right)\left(m-n^2\right)&=9\cdot1=3\cdot3 \end{aligned}\)
Karena m + n² > m − n², maka
\(\begin{aligned} m+n^2&=9\\ m-n^2&=1&\ +\\\hline 2m&=10\\ m&=5 \end{aligned}\)
Didapat, n = 2

b = 5³ = 125
d = 2⁵ = 32
125 − 32 = 93
9 + 3 = 12

\(\begin{aligned} x^3+(x+1)^3+\cdots+(x+7)^3&=y^3\\ 1^3+2^3+\cdots+(x+7)^3-\left(1^3+2^3+\cdots+(x-1)^3\right)^3&=y^3\\ (1+2+\cdots+x+7)^2-(1+2+\cdots+x-1)^2&=y^3\\ \left(\dfrac{(x+7)(x+8)}2\right)^2-\left(\dfrac{x(x-1)}2\right)^2&=y^3\\[4pt] \left(\dfrac{x^2+15x+56}2\right)^2-\left(\dfrac{x^2-x}2\right)^2&=y^3\\[4pt] \left(\dfrac{x^2+15x+56}2+\dfrac{x^2-x}2\right)\left(\dfrac{x^2+15x+56}2-\dfrac{x^2-x}2\right)&=y^3\\[4pt] \left(\dfrac{2x^2+14x+56}2\right)\left(\dfrac{16x+56}2\right)&=y^3\\[4pt] \left(x^2+7x+28\right)\left(8x+28\right)&=y^3\\[4pt] 4\left(x^2+7x+28\right)\left(2x+7\right)&=y^3 \end{aligned}\)
Misal 2x + 7 = t, dan y = nt
\(\begin{aligned} \left(t^2+63\right)t&=n^3t^3\\ t^2+63&=n^3t^2\\ \left(n^3-1\right)t^2&=63 \end{aligned}\)

• t² = 1
  t ∈ {−1,1}
  
  n³ = 64 ⟹ n = 4
  
  (x, y) ∈ {(−4, −4), (−3, 4)}
• t² = 9
  t ∈ {−3, 3}
  
  n³ = 8 ⟹ n = 2
  
  (x, y) ∈ {(−5, −6), (−2, 6)}

−4 − 4 − 3 + 4 − 5 − 6 − 2 + 6 = −14

Ketika a mendekati 1 dari kanan, nilai u akan semakin kecil dan mendekati −∞. Kita ubah batasnya yaitu $a,m\in\left(-1,-\dfrac12\right)\cup\left(\dfrac12,1\right)$.

\(\begin{aligned} 4a^2+m^2&\geq2\sqrt{4a^2m^2}\\ &\geq4 \end{aligned}\)

\(\begin{aligned} u&=\dfrac1{1-a^2}+\dfrac4{4-m^2}\\[4pt] &=\dfrac{4-m^2+4-4a^2}{4-4a^2-m^2+(am)^2}\\[4pt] &=\dfrac{8-4a^2-m^2}{5-4a^2-m^2}\\[4pt] &=1+\dfrac3{5-\left(4a^2+m^2\right)}\\[4pt] &\geq1+\dfrac3{5-4}=4 \end{aligned}\)

x + y = 5 − z

\(\begin{aligned} x^2+y^2+z^2&=(x+y+z)^2-2(xy+xz+yz)\\ &=5^2-2(3)\\ &=19\\ x^2+y^2&=19-z^2 \end{aligned}\)

dengan CSI didapat,
\(\begin{aligned} \left(x^2+y^2\right)\left(1^2+1^2\right)&\geq(x+y)^2\\ \left(19-z^2\right)\left(2\right)&\geq(5-z)^2\\ 38-2z^2&\geq25-10z+z^2\\ 3z^2-10z-13&\leq0\\ (3z-13)(z+1)&\leq0\\ \end{aligned}\)
$-1\leq z\leq\dfrac{13}3$
p = 13, q = 3
13 + 3 = 16